题目

Problem

A researcher is looking into the effectiveness of a new medicine for the relief of symptoms. He collects random samples of 8 people who are taking the medicine from each of 50 different medical practices. The number of people who say that the medicine is a success, in each sample, is recorded. The results are summarised in the table below.

Number of successes	0	1	2	3	4	5	6	7	8
Number of practices	4	6	3	12	10	7	4	2	2

The researcher decides to model this data using a binomial distribution.

(a) State two necessary assumptions that the researcher made in order to use this model.

(2)

(b) Show that the mean number of successes per sample is 3.54

(2)

He decides to use this mean to calculate expected frequencies. The results are shown in the table below.

Number of successes	0	1	2	3	4	5	6	7	8
Expected frequency	0.47	2.96	8.23	13.07	f	8.23	3.27	0.74	g

(4)

(d) Stating your hypotheses clearly, test at the 10% level of significance, whether or not the binomial distribution is a suitable model for the number of successes in samples of 8 people.

(8)

(Total for Question 5 is 16 marks)

题目中文翻译

一位研究者正在研究一种新药缓解症状的效果。他从 50 家不同的医疗机构中，每家随机抽取 8 名服药者组成样本，并记录每个样本中认为药物有效的人数。结果如下表。

成功次数	0	1	2	3	4	5	6	7	8
医疗机构数	4	6	3	12	10	7	4	2	2

研究者决定用二项分布来建模这些数据。

(a) 写出研究者为了使用这个模型所做的两条必要假设。

(b) 证明每个样本中的平均成功次数为 3.54。

他决定用这个均值来计算期望频数。结果如下表。

成功次数	0	1	2	3	4	5	6	7	8
期望频数	0.47	2.96	8.23	13.07	f	8.23	3.27	0.74	g

(d) 清楚写出假设，在 10% 显著性水平下检验：二项分布是否适合作为每组 8 人样本中成功人数的模型。

解答

(a)

解法一

思路

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二项分布要求固定次数的独立试验，并且每次成功概率相同。这里每家医疗机构的样本量已经固定为 $8$ ，所以需要明确说明不同服药者的结果相互独立，以及药物被认为有效的概率对所有服药者保持不变。

答题过程

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The necessary assumptions are:

Whether one person reports that the medicine is a success is independent of whether any other person reports success.
The probability that the medicine is reported as a success is the same for every person in every medical practice.

(b)

解法一

思路

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用每个成功次数乘以具有该成功次数的医疗机构数，求出 $50$ 家机构的成功总数，再除以机构数 $50$ 。题目要求 “Show that”，所以必须完整列出加权平均并自然得到 $3.54$ 。

答题过程

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The mean number of successes per sample is

\begin{align*} \bar{x} =&\,\frac{0(4)+1(6)+2(3)}{50}\\ &\,\hspace{2pt}+\frac{3(12)+4(10)+5(7)}{50}\\ &\,\hspace{4pt}+\frac{6(4)+7(2)+8(2)}{50}\\ =&\,\frac{177}{50}\\ =&\,\boxed{3.54}. \end{align*}

(c)

解法一

思路

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若每组成功人数 $X\sim\operatorname{Bin}(8,p)$ ，则 $\operatorname{E}(X)=8p$ 。用 (b) 的样本均值估计 $p$ ，再用“医疗机构数 $50$ 乘二项概率”分别计算成功 $4$ 次和 $8$ 次的期望频数。

答题过程

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Using the sample mean to estimate $p$ ,

\begin{align*} 8p =&\,3.54\\ p =&\,0.4425. \end{align*}

Let $X\sim\operatorname{Bin}(8,0.4425)$ . The expected frequency for four successes is

\begin{align*} f =&\,50\operatorname{P}(X=4)\\ =&\,50\binom{8}{4}\\ &\,\hspace{2pt}\times(0.4425)^4\\ &\,\hspace{4pt}\times(0.5575)^4\\ =&\,12.96285\ldots\\ =&\,\boxed{12.96}. \end{align*}

The expected frequency for eight successes is

\begin{align*} g =&\,50\operatorname{P}(X=8)\\ =&\,50(0.4425)^8\\ =&\,0.073498\ldots\\ =&\,\boxed{0.07}. \end{align*}

(d)

解法一

思路

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先提出二项模型是否合适的假设。原表有多个期望频数小于 $5$ ，因此把 $0,1,2$ 合并，并把 $5,6,7,8$ 合并，形成四组。参数 $p$ 由样本估计，所以自由度在“四组减一”之外还要再减一个估计参数，即 $4-1-1=2$ 。

答题过程

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The hypotheses are

\begin{aligned} H_0:&\ \text{a binomial distribution is a suitable model},\\ H_1:&\ \text{a binomial distribution is not a suitable model}. \end{aligned}

Classes with small expected frequencies are combined as follows.

Number of successes	Observed frequency, $O$	Expected frequency, $E$
$0,1,2$	13	11.6597
$3$	12	13.0654
$4$	10	12.9629
$5,6,7,8$	15	12.3120

The test statistic is

\begin{align*} \chi^2 =&\,\sum\frac{(O-E)^2}{E}\\ =&\,\frac{(13-11.6597)^2}{11.6597}\\ &\,\hspace{2pt}+\frac{(12-13.0654)^2}{13.0654}\\ &\,\hspace{4pt}+\frac{(10-12.9629)^2}{12.9629}\\ &\,\hspace{6pt}+\frac{(15-12.3120)^2}{12.3120}\\ =&\,1.50498\ldots\\ \approx&\,1.50. \end{align*}

There are four combined classes, and one parameter, $p$ , has been estimated from the data. Therefore,

\nu=4-1-1=2.

At the $10\%$ significance level, the critical value is

\chi_2^2(0.10)=4.605.

Since

1.50498\ldots<4.605,

the test statistic does not lie in the critical region, so $H_0$ is not rejected. There is insufficient evidence at the $10\%$ significance level to conclude that the binomial model is unsuitable. The data are consistent with a binomial distribution for the number of successes in samples of eight people.

IAL 2021 June S3 Q5

题目

解答

(a)

解法一

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(b)

解法一

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(c)

解法一

思路

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(d)

解法一

思路

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